23 Vectors

Learning Objectives

In this section you will:

  • View vectors geometrically.
  • Find magnitude and direction.
  • Perform vector addition and scalar multiplication.
  • Find the component form of a vector.
  • Find the unit vector in the direction of  \textbf{v}.
  • Perform operations with vectors in terms of  \textbf{i} and  \textbf{j}.
  • Find the dot product of two vectors.

An airplane is flying at an airspeed of 200 miles per hour headed on a SE bearing of 140°. A north wind (from north to south) is blowing at 16.2 miles per hour, as shown in (Figure 1). What are the ground speed and actual bearing of the plane?

Image of a plan flying SE at 140 degrees and the north wind blowing
Figure 1.

Ground speed refers to the speed of a plane relative to the ground. Airspeed refers to the speed a plane can travel relative to its surrounding air mass. These two quantities are not the same because of the effect of wind. In an earlier section, we used triangles to solve a similar problem involving the movement of boats. Later in this section, we will find the airplane’s groundspeed and bearing, while investigating another approach to problems of this type. First, however, let’s examine the basics of vectors.

A Geometric View of Vectors

A vector is a specific quantity drawn as a line segment with an arrowhead at one end. It has an initial point, where it begins, and a terminal point, where it ends. A vector is defined by its magnitude, or the length of the line, and its direction, indicated by an arrowhead at the terminal point. Thus, a vector is a directed line segment. There are various symbols that distinguish vectors from other quantities:

  • Lower case, boldfaced type, with or without an arrow on top such as \textbf {v},\,\,\textbf {u},\,\,\textbf {w},\,\,\stackrel{\to }{v},\,\,\stackrel{\to }{u},\,\stackrel{\to }{w}.
  • Given initial point \,P\, and terminal point \,Q,\, a vector can be represented as \,\,\stackrel{\to }{PQ}\,.\,\, The arrowhead on top is what indicates that it is not just a line, but a directed line segment.
  • Given an initial point of \,\left(0,0\right)\, and terminal point \,\left(a,b\right),\, a vector may be represented as \,\big\langle a,b\big\rangle.

This last symbol 〈a,b〉 has special significance. It is called the standard position. The position vector has an initial point \left(0,0\right)\, and a terminal point 〈a,b〉. To change any vector into the position vector, we think about the change in the x-coordinates and the change in the y-coordinates. Thus, if the initial point of a vector \,\stackrel{\to }{CD}\, is \,C\left({x}_{1},{y}_{1}\right)\, and the terminal point is \,D\left({x}_{2},{y}_{2}\right),\, then the position vector is found by calculating

\begin{array}{l}\stackrel{\to }{AB}\,=\,〈{x}_{2}-{x}_{1},{y}_{2}-{y}_{1}〉\hfill \\ \,\,\,\,\,\,\,\,\,\,=\,\text\,\big\langle a,b\big\rangle\hfill \end{array}

In (Figure 2), we see the original vector \,\stackrel{\to }{CD}\, and the position vector\,\stackrel{\to }{AB}.

Plot of the original vector CD in blue and the position vector AB in orange extending from the origin.
Figure 2.

Properties of Vectors

A vector is a directed line segment with an initial point and a terminal point. Vectors are identified by magnitude, or the length of the line, and direction, represented by the arrowhead pointing toward the terminal point. The position vector has an initial point at \,\left(0,0\right)\, and is identified by its terminal point \big\langle a,b\big\rangle.

Find the Position Vector

Consider the vector whose initial point is \,P\left(2,3\right)\,and terminal point is\,Q\left(6,4\right).\,Find the position vector.

Show Solution

The position vector is found by subtracting one x-coordinate from the other x-coordinate, and one y-coordinate from the other y-coordinate. Thus

\begin{array}{l}\textbf{v}=〈6-2,4-3〉\hfill \\ \,\,\,=\big\langle 4,1\big\rangle\hfill \end{array}

The position vector begins at \,\left(0,0\right)\, and terminates at \,\left(4,1)\right.\, The graphs of both vectors are shown in (Figure 3).

Plot of the original vector in blue and the position vector in orange extending from the origin.
Figure 3.

We see that the position vector is \big\langle 4,1\big\rangle.

Drawing a Vector with the Given Criteria and Its Equivalent Position Vector

Find the position vector given that vector \, \textbf{v}\, has an initial point at \,\left(-3,2\right)\,and a terminal point at \,\left(4,5\right),\, then graph both vectors in the same plane.

Show Solution

The position vector is found using the following calculation:

\begin{array}{l}\textbf{v}=〈4-\left(-3\right),5-2〉\hfill \\ \big\langle 7,3\big\rangle\hfill \end{array}

Thus, the position vector begins at \,\left(0,0\right)\, and terminates at \,\left(7,3\right).\, See (Figure 4).

Plot of the two given vectors their same position vector.
Figure 4.

Try It

Draw a vector \,\textbf{v}\, that connects from the origin to the point \,\left(3,5\right).

Show Solution

A vector from the origin to (3,5) - a line with an arrow at the (3,5) endpoint.

Finding Magnitude and Direction

To work with a vector, we need to be able to find its magnitude and its direction. We find its magnitude using the Pythagorean Theorem or the distance formula, and we find its direction using the inverse tangent function.

Magnitude and Direction of a Vector

Given a position vector \,\textbf{v}=\big\langle a,b\big\rangle, the magnitude is found by |v|=\sqrt{{a}^{2}+{b}^{2}}. The direction is equal to the angle formed with the x-axis, or with the y-axis, depending on the application. For a position vector, the direction is found by \,\mathrm{tan}\,\theta =\left(\frac{b}{a}\right);⇒\,\,\theta ={\mathrm{tan}}^{-1}\left(\frac{b}{a}\right),\, as illustrated in (Figure 5).

Standard plot of a position vector (a,b) with magnitude |v| extending into Q1 at theta degrees.
Figure 5.

Two vectors v and u are considered equal if they have the same magnitude and the same direction. Additionally, if both vectors have the same position vector, they are equal.

Finding the Magnitude and Direction of a Vector

Find the magnitude and direction of the vector with initial point \,P\left(-8,1\right)\, and terminal point \,Q\left(-2,-5\right). Draw the vector.

Show Solution

First, find the position vector.

\begin{array}{l}\textbf{u}=〈-2,-\left(-8\right),-5-1〉\hfill \\ \big\langle 6,-6\big\rangle\hfill \end{array}

We use the Pythagorean Theorem to find the magnitude.

\begin{array}{l}|u|=\sqrt{{\left(6\right)}^{2}+{\left(-6\right)}^{2}}\hfill \\ \,\,\,\,\,\,=\sqrt{72}\hfill \\ \,\,\,\,\,\,=6\sqrt{2}\hfill \end{array}

The direction is given as

\begin{array}{l}\mathrm{tan}\,\theta =\frac{-6}{6}=-1;\,\,\,⇒\theta ={\mathrm{tan}}^{-1}\left(-1\right)\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,=-45^{\circ }\hfill \end{array}

However, the angle terminates in the fourth quadrant, so we add 360° to obtain a positive angle. Thus,\,-45^{\circ }+360^{\circ }=315^{\circ }.\, See (Figure 6).

Plot of the position vector extending into Q4 from the origin with the magnitude 6rad2.
Figure 6.

Showing That Two Vectors Are Equal

Show that vector v with initial point at \,\left(5,-3\right)\, and terminal point at \,\left(-1,2\right)\, is equal to vector u with initial point at \,\left(-1,-3\right)\, and terminal point at \,\left(-7,2\right).\, Draw the position vector on the same grid as v and u. Next, find the magnitude and direction of each vector.

Show Solution

As shown in (Figure 7), draw the vector \,\textbf{v}\, starting at initial \,\left(5,-3\right)\, and terminal point \,\left(-1,2\right).\, Draw the vector \,\textbf{u}\, with initial point\,\left(-1,-3\right)\, and terminal point \,\left(-7,2\right).\, Find the standard position for each.

Next, find and sketch the position vector for v and u. We have

\begin{array}{l}\textbf{v}=〈-1-5,2-\left(-3\right)〉\hfill \\ \big\langle -6,5\big\rangle\hfill \\ \hfill \\ \textbf{u}=〈-7-\left(-1\right),2-\left(-3\right)〉\hfill \\ \big\langle -6,5\big\rangle\hfill \end{array}

Since the position vectors are the same, v and u are the same.

An alternative way to check for vector equality is to show that the magnitude and direction are the same for both vectors. To show that the magnitudes are equal, use the Pythagorean Theorem.

\begin{array}{l}|v|=\sqrt{{\left(-1-5\right)}^{2}+{\left(2-\left(-3\right)\right)}^{2}}\hfill \\ \,\,\,\,\,\,=\sqrt{{\left(-6\right)}^{2}+{\left(5\right)}^{2}}\hfill \\ \,\,\,\,\,\,=\sqrt{36+25}\hfill \\ \,\,\,\,\,\,=\sqrt{61}\hfill \\ |u|=\sqrt{{\left(-7-\left(-1\right)\right)}^{2}+{\left(2-\left(-3\right)\right)}^{2}}\hfill \\ \,\,\,\,\,\,=\sqrt{{\left(-6\right)}^{2}+{\left(5\right)}^{2}}\hfill \\ \,\,\,\,\,\,=\sqrt{36+25}\hfill \\ \,\,\,\,\,\,=\sqrt{61}\hfill \end{array}

As the magnitudes are equal, we now need to verify the direction. Using the tangent function with the position vector gives

\begin{array}{l}\mathrm{tan}\,\theta =-\frac{5}{6};\,\,\,⇒\theta ={\mathrm{tan}}^{-1}\left(-\frac{5}{6}\right)\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\theta =-39.8^{\circ }\hfill \end{array}

However, we can see that the position vector terminates in the second quadrant, so we add \,180^{\circ }.\, Thus, the direction is \,-39.8^{\circ }+180^{\circ }=140.2^{\circ }.

Plot of the two given vectors their same position vector.
Figure 7.

Performing Vector Addition and Scalar Multiplication

Now that we understand the properties of vectors, we can perform operations involving them. While it is convenient to think of the vector  \textbf{u}=〈x,y〉 as an arrow or directed line segment from the origin to the point \,\left(x,y\right),\, vectors can be situated anywhere in the plane. The sum of two vectors u and v, or vector addition, produces a third vector u+ v, the resultant vector.

To find u + v, we first draw the vector u, and from the terminal end of u, we drawn the vector v. In other words, we have the initial point of v meet the terminal end of u. This position corresponds to the notion that we move along the first vector and then, from its terminal point, we move along the second vector. The sum u + v is the resultant vector because it results from addition or subtraction of two vectors. The resultant vector travels directly from the beginning of u to the end of v in a straight path, as shown in (Figure 8).

Diagrams of vector addition and subtraction.
Figure 8.

Vector subtraction is similar to vector addition. To find u − v, view it as u + (−v). Adding − v is reversing direction of v and adding it to the end of u. The new vector begins at the start of u and stops at the end point of − v. See (Figure 9) for a visual that compares vector addition and vector subtraction using parallelograms.

Showing vector addition and subtraction with parallelograms. For addition, the base is u, the side is v, the diagonal connecting the start of the base to the end of the side is u+v. For subtraction, thetop is u, the side is -v, and the diagonal connecting the start of the top to the end of the side is u-v.
Figure 9.

Adding and Subtracting Vectors

Given \textbf{u}=\big\langle 3,-2\big\rangle and \textbf{v}=\big\langle -1,4\big\rangle, find two new vectors u + v, and u − v.

Show Solution

To find the sum of two vectors, we add the components. Thus,

\begin{array}{l}\textbf{u}+\textbf{v}=\big\langle 3,-2\big\rangle+\big\langle -1,4\big\rangle\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,=〈3+\left(-1\right),-2+4〉\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,=\big\langle 2,2\big\rangle\hfill \end{array}

See (Figure 10)(a).

To find the difference of two vectors, add the negative components of \,\textbf{v}\, to \,\textbf{u}.\, Thus,

\begin{array}{l}\textbf{u}+\left(-\textbf{v}\right)=\big\langle 3,-2\big\rangle+\big\langle 1,-4\big\rangle\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=〈3+1,-2+\left(-4\right)〉\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\big\langle 4,-6\big\rangle\hfill \end{array}

See (Figure 10)(b).

Further diagrams of vector addition and subtraction.
Figure 10. (a) Sum of two vectors (b) Difference of two vectors

Multiplying By a Scalar

While adding and subtracting vectors gives us a new vector with a different magnitude and direction, the process of multiplying a vector by a scalar, a constant, changes only the magnitude of the vector or the length of the line. Scalar multiplication has no effect on the direction unless the scalar is negative, in which case the direction of the resulting vector is opposite the direction of the original vector.

Scalar Multiplication

Scalar multiplication involves the product of a vector and a scalar. Each component of the vector is multiplied by the scalar. Thus, to multiply \textbf{v}=〈a,b〉 by k, we have

k\textbf{v}=〈ka,kb〉

Only the magnitude changes, unless \,k\, is negative, and then the vector reverses direction.

Performing Scalar Multiplication

Given vector \,\textbf{v}=\big\langle 3,1\big\rangle,\, find 3v, \frac{1}{2}\textbf{v},\, and − v.

Show Solution

See (Figure 11) for a geometric interpretation. If \,\textbf{v}=\big\langle 3,11\big\rangle, then

\begin{array}{l}\,\,3\textbf{v}=〈3\cdot 3,3\cdot 1〉\hfill \\ \,\,\,\,\,\,\,\,=\big\langle 9,3\big\rangle\hfill \\ \,\frac{1}{2}\textbf{v}=\frac{1}{2}\cdot 3,\frac{1}{2}\cdot 1〉\hfill \\ \,\,\,\,\,\,\,\,=〈\big\langle\frac{3}{2},\frac{1}{2}〉\big\rangle\hfill \\ -\textbf{v}=\big\langle -3,-1\big\rangle\hfill \end{array}
Showing the effect of scaling a vector: 3x, 1x, .5x, and -1x. The 3x is three times as long, the 1x stays the same, the .5x halves the length, and the -1x reverses the direction of the vector but keeps the length the same. The rest keep the same direction; only the magnitude changes.
Figure 11.

Analysis

Notice that the vector 3v is three times the length of v, \frac{1}{2} \textbf{v}\, is half the length of v, and – v is the same length of v, but in the opposite direction.

Try It

Find the scalar multiple 3\textbf{u} given \textbf{u}=〈5,4〉.

Show Solution

3\textbf{u}=\big\langle 15,12\big\rangle

Using Vector Addition and Scalar Multiplication to Find a New Vector

Given \textbf{u} =\big\langle 3,-2\big\rangle/latex] and <strong>[latex]\textbf{v}=\big\langle -1,4\big\rangle, find a new vector w = 3u + 2v.

Show Solution

First, we must multiply each vector by the scalar.

\begin{array}{l}3\textbf{u}=3\cdot 3,3\cdot -2〉\hfill \\ \,\,\,\,\,\,=\big\langle 9,-6\big\rangle\hfill \\ 2\textbf{v}=2\cdot-1, 2\cdot 4〉\hfill \\ \,\,\,\,\,\,=\big\langle -2,8\big\rangle\hfill \end{array}

Then, add the two together.

\begin{array}{l}\textbf{w}=3\textbf{u}+2\textbf{v}\hfill \\ \,\,\,\,\,=\big\langle 9,-6\big\rangle+\big\langle -2,8\big\rangle\hfill \\ \,\,\,\,\,=\big\langle 9-2,-6+8\big\rangle\hfill \\ \,\,\,\,\,=\big\langle 7,2\big\rangle\hfill \end{array}

So, \textbf{w}=\big\langle 7,2\big\rangle.

Finding Component Form

In some applications involving vectors, it is helpful for us to be able to break a vector down into its components. Vectors are comprised of two components: the horizontal component is the\,x\,direction, and the vertical component is the\,y\,direction. For example, we can see in the graph in (Figure 12) that the position vector \big\langle 2,3\big\rangle comes from adding the vectors v1 and v2. We have v1 with initial point\,\left(0,0\right)\,and terminal point \,\left(2,0\right).\,

\begin{array}{l}\textbf{v}_{1}=\big\langle 2-0,0-0\big\rangle\hfill \\ \,\,\,\,\,\,=\big\langle 2,0\big\rangle\hfill \end{array}

We also have v2 with initial poin t\,\left(0,0\right)\, and terminal point\,\left(0,\,3\right).\,

\begin{array}{l}\textbf{v}_{2}=\big\langle 0-0,3-0\big\rangle\hfill \\ \,\,\,\,\,\,\,=\big\langle 0,3\big\rangle\hfill \end{array}

Therefore, the position vector is

\begin{array}{l}\textbf{v}=\big\langle 2+0,3+0\big\rangle\hfill \\ \,\,\,=\big\langle 2,3\big\rangle\hfill \end{array}

Using the Pythagorean Theorem, the magnitude of v1 is 2, and the magnitude of v2 is 3. To find the magnitude of v, use the formula with the position vector.

\begin{array}{l}\hfill \\ \begin{array}{l}|v|=\sqrt{|{v}_{1}{|}^{2}+|{v}_{2}{|}^{2}}\hfill \\ \begin{array}{l}\,\,\,\,\,=\sqrt{{2}^{2}+{3}^{2}}\hfill \\ \,\,\,\,\,=\sqrt{13}\hfill \end{array}\hfill \end{array}\hfill \end{array}

The magnitude of v is \,\sqrt{13}.\, To find the direction, we use the tangent function \,\mathrm{tan}\,\theta =\frac{y}{x}.

\begin{array}{l}\mathrm{tan}\,\theta =\frac{{v}_{2}}{{v}_{1}}\hfill \\ \mathrm{tan}\,\theta =\frac{3}{2}\hfill \\ \,\,\,\,\,\,\,\,\,\theta ={\mathrm{tan}}^{-1}\left(\frac{3}{2}\right)=56.3^{\circ }\hfill \end{array}
Diagram of a vector in root position with its horizontal and vertical components.
Figure 12.

Thus, the magnitude of \,\textbf{v}\, is \,\sqrt{13}\, and the direction is \,{56.3}^{\circ } off the horizontal.

Finding the Components of the Vector

Find the components of the vector \,\textbf{v}\, with initial point \,\left(3,2\right)\, and terminal point\,\left(7,4\right).

Show Solution

First find the standard position.

\begin{array}{l}\textbf{v}=\big\langle 7-3,4-2\big\rangle\hfill \\ \,\,\,=\big\langle 4,2\big\rangle\hfill \end{array}

See the illustration in (Figure 13).

Diagram of a vector in root position with its horizontal (4,0) and vertical (0,2) components.
Figure 13.

The horizontal component is \textbf{v}_{1}=\big\langle 4,0\big\rangle\, and the vertical component is \,\textbf{v}_{2}=\big\langle 0,2\big\rangle.

Finding the Unit Vector in the Direction of v

In addition to finding a vector’s components, it is also useful in solving problems to find a vector in the same direction as the given vector, but of magnitude 1. We call a vector with a magnitude of 1 a unit vector. We can then preserve the direction of the original vector while simplifying calculations.

Unit vectors are defined in terms of components. The horizontal unit vector is written as \textbf{i}=\big\langle 1,0\big\rangle and is directed along the positive horizontal axis. The vertical unit vector is written as \textbf{j}=\big\langle 0,1\big\rangle and is directed along the positive vertical axis. See (Figure 14).

Plot showing the unit vectors i=91,0) and j=(0,1)
Figure 14.

The Unit Vectors

If \,\textbf{v}\, is a nonzero vector, then \,\frac{v}{|v|}\, is a unit vector in the direction of \,\textbf{v}.\, Any vector divided by its magnitude is a unit vector. Notice that magnitude is always a scalar, and dividing by a scalar is the same as multiplying by the reciprocal of the scalar.

Finding the Unit Vector in the Direction of v

Find a unit vector in the same direction as \textbf{v}=\big\langle -5,12\big\rangle.

Show Solution

First, we will find the magnitude.

\begin{array}{l}|v|=\sqrt{{\left(-5\right)}^{2}+{\left(12\right)}^{2}}\hfill \\ \,\,\,\,\,\,=\sqrt{25+144}\hfill \\ \,\,\,\,\,\,=\sqrt{169}\hfill \\ \,\,\,\,\,\,=13\hfill \end{array}

Then we divide each component by \,|v|,\, which gives a unit vector in the same direction as v:

\frac{v}{|v|}=-\frac{5}{13}\textbf{i}}+\frac{12}{13}\textbf{j}}

or, in component form

\frac{v}{|v|}=\big\langle-\frac{5}{13},\frac{12}{13}\big\rangle

See (Figure 15).

Plot showing the unit vector (-5/13, 12/13) in the direction of (-5, 12)
Figure 15.

Verify that the magnitude of the unit vector equals 1. The magnitude of\,-\frac{5}{13}\textbf{i}+\frac{12}{13}\textbf{j}\, is given as

\begin{array}{l}\sqrt{{\left(-\frac{5}{13}\right)}^{2}+{\left(\frac{12}{13}\right)}^{2}}=\sqrt{\frac{25}{169}+\frac{144}{169}}\hfill \\ \text{ }=\sqrt{\frac{169}{169}}=1\hfill \end{array}

The vector u =\frac{5}{13} i +\frac{12}{13} j is the unit vector in the same direction as v =\big\langle -5,12\big\rangle.

Performing Operations with Vectors in Terms of i and j

So far, we have investigated the basics of vectors: magnitude and direction, vector addition and subtraction, scalar multiplication, the components of vectors, and the representation of vectors geometrically. Now that we are familiar with the general strategies used in working with vectors, we will represent vectors in rectangular coordinates in terms of i and j.

Vectors in the Rectangular Plane

Given a vector \,\textbf{v}\, with initial point \,P=\left({x}_{1},{y}_{1}\right)\,and terminal point Q=\left({x}_{2},{y}_{2}\right), v is written as

\textbf{v}=\left({x}_{2}-{x}_{1}\right)\textbf{i}+\left({y}_{2}-{y}_{1}\right)\textbf{j}

The position vector from \,\left(0,0\right)\, to \,\left(a,b\right),\, where \,\left({x}_{2}-{x}_{1}\right)=a\, and \,\left({y}_{2}-{y}_{1}\right)=b,\, is written as v = ai + bj. This vector sum is called a linear combination of the vectors i and j.

The magnitude of v = ai + bj is given as \,|v|=\sqrt{{a}^{2}+{b}^{2}}.\, See (Figure 16).

Plot showing vectors in rectangular coordinates in terms of i and j. The position vector v (in orange) extends from the origin to some point (a,b) in Q1. The horizontal (ai) and vertical (bj) components are shown.
Figure 16.

Writing a Vector in Terms of i and j

Given a vector \,\textbf{v}\, with initial point \,P=\left(2,-6\right)\, and terminal point \,Q=\left(-6,6\right),\, write the vector in terms of \,\textbf{i}\, and \,\textbf{j}.

Show Solution

Begin by writing the general form of the vector. Then replace the coordinates with the given values.

\begin{array}{l}\textbf{v}=\left({x}_{2}-{x}_{1}\right)\textbf{i}+\left({y}_{2}-{y}_{1}\right)\textbf{j}\hfill \\ \,\,\,=\left(-6-2\right)\textbf{i}+\left(6-\left(-6\right)\right)\textbf{j}\hfill \\ \,\,\,=-8\textbf{i}+12\textbf{j}\hfill \end{array}

Writing a Vector in Terms of i and j Using Initial and Terminal Points

Given initial point \,{P}_{1}=\left(-1,3\right)\, and terminal point \,{P}_{2}=\left(2,7\right),\, write the vector \,\textbf{v}\, in terms of \,\textbf{i}\, and \,\textbf{j}.\,

Show Solution

Begin by writing the general form of the vector. Then replace the coordinates with the given values.

\begin{array}{l}\textbf{v}=\left({x}_{2}-{x}_{1}\right)\textbf{i}+\left({y}_{2}-{y}_{1}\right)\textbf{j}\hfill \\ v=\left(2-\left(-1\right)\right)\textbf{i}+\left(7-3\right)\textbf{j}\hfill \\ \,\,=3\textbf{i}+4\textbf{j}\hfill \end{array}

Try It

Write the vector \,\textbf{u}\, with initial point \,P=\left(-1,6\right)\, and terminal point \,Q=\left(7,-5\right)\, in terms of \,\textbf{i}\, and \,\textbf{j}.

Show Solution

\textbf{u}=8\textbf{i}-11\textbf{j}

Performing Operations on Vectors in Terms of i and j

When vectors are written in terms of \,\textbf{i}\, and \,\textbf{j},\, we can carry out addition, subtraction, and scalar multiplication by performing operations on corresponding components.

Adding and Subtracting Vectors in Rectangular Coordinates

Given v = ai + bj and u = ci + dj, then

\begin{array}{c}\textbf{v}+\textbf{u}=\left(a+c\right)\textbf{i}+\left(b+d\right)\textbf{j}\\ \textbf{v}-\textbf{u}=\left(a-c\right)\textbf{i}+\left(b-d\right)\textbf{j}\end{array}

Finding the Sum of the Vectors

Find the sum of \,\textbf{v}_{1}=2\textbf{i}-3\textbf{j}\, and \,\textbf{v}_{2}=4\textbf{i}+5\textbf{j}.

Show Solution

According to the formula, we have

\begin{array}{l}\textbf{v}_{1}+\textbf{v}_{2}=\left(2+4\right)\textbf{i}+\left(-3+5\right)\textbf{j}\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=6\textbf{i}+2\textbf{j}\hfill \end{array}

Calculating the Component Form of a Vector: Direction

We have seen how to draw vectors according to their initial and terminal points and how to find the position vector. We have also examined notation for vectors drawn specifically in the Cartesian coordinate plane using \,\textbf{i}\,\,\text{and}\,\,\textbf{j}.\, For any of these vectors, we can calculate the magnitude. Now, we want to combine the key points, and look further at the ideas of magnitude and direction.

Calculating direction follows the same straightforward process we used for polar coordinates. We find the direction of the vector by finding the angle to the horizontal. We do this by using the basic trigonometric identities, but with \,|v|\, replacing \,r.

Vector Components in Terms of Magnitude and Direction

Given a position vector \,\textbf{v}=〈x,y〉\, and a direction angle \,\theta ,

\begin{array}{lll}\mathrm{cos}\,\theta =\frac{x}{|v|}\hfill & \text{and}\begin{array}{cc}& \end{array}\hfill & \mathrm{sin}\,\theta =\frac{y}{|v|}\hfill \\ \,\,\,\,\,\,\,\,x=|v|\mathrm{cos}\,\theta \begin{array}{cc}& \end{array}\hfill & \hfill & \,\,\,\,\,\,\,y=|v|\mathrm{sin}\,\theta \hfill \end{array}

Thus, \,\textbf{v}=x\textbf{i}+y\textbf{j}=|v|\mathrm{cos}\,\theta \textbf{i}+|v|\mathrm{sin}\,\theta \textbf{j},\, and magnitude is expressed as \,|v|=\sqrt{{x}^{2}+{y}^{2}}.

Writing a Vector in Terms of Magnitude and Direction

Write a vector with length 7 at an angle of 135° to the positive
x-axis in terms of magnitude and direction.

Show Solution

Using the conversion formulas \, x = |v|\mathrm{cos}\,\theta \textbf{i}\, and \,y=|v|\mathrm{sin}\,\theta \textbf{j},\, we find that

\begin{array}{l}x=7\mathrm{cos}\left(135^{\circ }\right)\textbf{i}\hfill \\ \,\,\,=-\frac{7\sqrt{2}}{2}\hfill \\ y=7\mathrm{sin}\left(135^{\circ }\right)\textbf{j}\hfill \\ \,\,\,=\frac{7\sqrt{2}}{2}\hfill \end{array}

This vector can be written as \,\textbf{v}=7\mathrm{cos}\left(135^{\circ }\right)\textbf{i}+7\mathrm{sin}\left(135^{\circ }\right)\textbf{j}\, or simplified as

\textbf{v}=-\frac{7\sqrt{2}}{2}\textbf{i}+\frac{7\sqrt{2}}{2}\textbf{j}

A vector travels from the origin to the point \,\left(3,5\right).\, Write the vector in terms of magnitude and direction.

Show Solution

\textbf{v}=\sqrt{34}\mathrm{cos}\left(59^{\circ }\right)\textbf{i}+\sqrt{34}\mathrm{sin}\left(59^{\circ }\right)\textbf{j}

Magnitude =\,\sqrt{34}

\theta ={\mathrm{tan}}^{-1}\left(\frac{5}{3}\right)=59.04^{\circ }

Finding the Dot Product of Two Vectors

As we discussed earlier in the section, scalar multiplication involves multiplying a vector by a scalar, and the result is a vector. As we have seen, multiplying a vector by a number is called scalar multiplication. If we multiply a vector by a vector, there are two possibilities: the dot product and the cross product. We will only examine the dot product here; you may encounter the cross product in more advanced mathematics courses.

The dot product of two vectors involves multiplying two vectors together, and the result is a scalar.

Dot Product

The dot product of two vectors \,\textbf{v}=〈a,b〉\, and \,\textbf{v}=〈c,d〉\, is the sum of the product of the horizontal components and the product of the vertical components.

\textbf{v}\cdot \textbf{u}=ac+bd

To find the angle between the two vectors, use the formula below.

\mathrm{cos}\,\theta =\frac{v}{|v|}\cdot \frac{u}{|u|}

Finding the Dot Product of Two Vectors

Find the dot product of \,\textbf{v}=\big\langle 5,12\big\rangle\, and \,\textbf{u}=\big\langle -3,4\big\rangle.

Show Solution

Using the formula, we have

\begin{array}{l}\textbf{v}\cdot \textbf{v}=\big\langle 5,12\big\rangle\cdot \big\langle -3,4\big\rangle\hfill \\ \,\,\,\,\,\,\,\,\,\,=5\cdot \left(-3\right)+12\cdot 4\hfill \\ \,\,\,\,\,\,\,\,\,\,=-15+48\hfill \\ \,\,\,\,\,\,\,\,\,\,=33\hfill \end{array}

Finding the Dot Product of Two Vectors and the Angle between Them

Find the dot product of v1 = 5i + 2j and v2 = 3i + 7j. Then, find the angle between the two vectors.

Show Solution

Finding the dot product, we multiply corresponding components.

\begin{array}{l}\textbf{v}_{1}\cdot \textbf{v}_{2}=\big\langle 5,2\big\rangle\cdot \big\langle 3,7\big\rangle\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=5\cdot 3+2\cdot 7\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=15+14\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=29\hfill \end{array}

To find the angle between them, we use the formula\,\mathrm{cos}\,\theta =\frac\textbf{v}{|v|}\cdot \frac\textbf{v}{|u|}.

\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{v}{|v|}\cdot \frac{u}{|u|}=〈\frac{5}{\sqrt{29}}+\frac{2}{\sqrt{29}}〉\cdot 〈\frac{3}{\sqrt{58}}+\frac{7}{\sqrt{58}}〉\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{5}{\sqrt{29}}\cdot \frac{3}{\sqrt{58}}+\frac{2}{\sqrt{29}}\cdot \frac{7}{\sqrt{58}}\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{15}{\sqrt{1682}}+\frac{14}{\sqrt{1682}}=\frac{29}{\sqrt{1682}}\hfill \\ \begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=0.707107\hfill \\ {\mathrm{cos}}^{-1}\left(0.707107\right)=45^{\circ }\hfill \end{array}\hfill \end{array}

See (Figure 17).

Plot showing the two position vectors (3,7) and (5,2) and the 45 degree angle between them.
Figure 17.

Finding the Angle between Two Vectors

Find the angle between \,\textbf{v}=\big\langle -3,4\big\rangle\, and \,\textbf{u}=\big\langle 5,12\big\rangle.

Show Solution

Using the formula, we have

\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\theta ={\mathrm{cos}}^{-1}\left(\frac{u}{|u|}\cdot \frac{v}{|v|}\right)\hfill \\ \left(\frac{u}{|u|}\cdot \frac{v}{|v|}\right)=\frac{-3i+4j}{5}\cdot \frac{5i+12j}{13}\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left(-\frac{3}{5}\cdot \frac{5}{13}\right)+\left(\frac{4}{5}\cdot \frac{12}{13}\right)\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=-\frac{15}{65}+\frac{48}{65}\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{33}{65}\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\theta ={\mathrm{cos}}^{-1}\left(\frac{33}{65}\right)\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,={59.5}^{\circ }\hfill \end{array}

See (Figure 18).

Plot showing the two position vectors (-3,4) and (5,12) and the 59.5 degree angle between them.
Figure 18.

Finding Ground Speed and Bearing Using Vectors

We now have the tools to solve the problem we introduced in the opening of the section.

An airplane is flying at an airspeed of 200 miles per hour headed on a SE bearing of 140°. A north wind (from north to south) is blowing at 16.2 miles per hour. What are the ground speed and actual bearing of the plane? See (Figure 19).

Image of a plan flying SE at 140 degrees and the north wind blowing.
Figure 19.
Show Solution

The ground speed is represented by \,x\, in the diagram, and we need to find the angle \,\alpha \, in order to calculate the adjusted bearing, which will be \,\,140^{\circ }+\alpha \,.

Notice in (Figure19), that angle \,BCO\, must be equal to angle \,AOC\, by the rule of alternating interior angles, so angle \,BCO\, is 140^{\circ }. We can find \,x\, by the Law of Cosines:

\begin{array}{l}{x}^{2}={\left(16.2\right)}^{2}+{\left(200\right)}^{2}-2\left(16.2\right)\left(200\right)\mathrm{cos}\left(140^{\circ }\right)\hfill \\ {x}^{2}=45,226.41\hfill \\ \,\,\,x=\sqrt{45,226.41}\hfill \\ \,\,\,x=212.7\hfill \end{array}

The ground speed is approximately 213 miles per hour. Now we can calculate the bearing using the Law of Sines.

\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{\mathrm{sin}\,\alpha }{16.2}=\frac{\mathrm{sin}\left(140^{\circ }\right)}{212.7}\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\mathrm{sin}\,\alpha =\frac{16.2\mathrm{sin}\left(140^{\circ }\right)}{212.7}\hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=0.04896\hfill \\ {\mathrm{sin}}^{-1}\left(0.04896\right)=2.8^{\circ }\hfill \end{array}

Therefore, the plane has a SE bearing of 140°+2.8°=142.8°. The ground speed is 212.7 miles per hour.

Access these online resources for additional instruction and practice with vectors.

Key Concepts

Section Exercises

Verbal

What are the characteristics of the letters that are commonly used to represent vectors?

Show Solution

lowercase, bold letter, usually \,u,v,w

How is a vector more specific than a line segment?

What are\,\textbf{i}\,and\,\textbf{j},and what do they represent?

Show Solution

They are unit vectors. They are used to represent the horizontal and vertical components of a vector. They each have a magnitude of 1.

What is component form?

When a unit vector is expressed as 〈a,b〉,which letter is the coefficient of the \,\textbf{i}\, and which the \,\textbf{j}?

Show Solution

The first number always represents the coefficient of the\,i,\,and the second represents the\,j.

Algebraic

Given a vector with initial point \,\left(5,2\right)\, and terminal point \,\left(-1,-3\right),\, find an equivalent vector whose initial point is \,\left(0,0\right).\,Write the vector in component form \big\langle a,b\big\rangle.

Given a vector with initial point \,\left(-4,2\right)\, and terminal point \,\left(3,-3\right),\, find an equivalent vector whose initial point is \,\left(0,0\right).\, Write the vector in component form \big\langle a,b\big\rangle.

Show Solution

\big\langle -7,5\big\rangle〉

Given a vector with initial point \,\left(7,-1\right)\, and terminal point \,\left(-1,-7\right),\, find an equivalent vector whose initial point is \,\left(0,0\right).\, Write the vector in component form \big\langle a,b\big\rangle.

For the following exercises, determine whether the two vectors \,\textbf{u}\, and \,\textbf{v}\, are equal, where \,\textbf{u}\, has an initial point \,{P}_{1}\, and a terminal point \,{P}_{2}\, and \textbf{v} has an initial point \,{P}_{3}\, and a terminal point \,{P}_{4}.

{P}_{1}=\left(5,1\right),{P}_{2}=\left(3,-2\right),{P}_{3}=\left(-1,3\right),\, and \,{P}_{4}=\left(9,-4\right)

Show Solution

not equal

{P}_{1}=\left(2,-3\right),{P}_{2}=\left(5,1\right),{P}_{3}=\left(6,-1\right),\, and \,{P}_{4}=\left(9,3\right)

{P}_{1}=\left(-1,-1\right),{P}_{2}=\left(-4,5\right),{P}_{3}=\left(-10,6\right),\, and \,{P}_{4}=\left(-13,12\right)

Show Solution

equal

{P}_{1}=\left(3,7\right),{P}_{2}=\left(2,1\right),{P}_{3}=\left(1,2\right),\, and \,{P}_{4}=\left(-1,-4\right)

{P}_{1}=\left(8,3\right),{P}_{2}=\left(6,5\right),{P}_{3}=\left(11,8\right),\, and {P}_{4}=\left(9,10\right)

Show Solution

equal

Given initial point \,{P}_{1}=\left(-3,1\right)\, and terminal point \,{P}_{2}=\left(5,2\right),\, write the vector \,\textbf{v}\, in terms of \,\textbf{i}\, and \,\textbf{j}.\,

Given initial point \,{P}_{1}=\left(6,0\right)\, and terminal point \,{P}_{2}=\left(-1,-3\right),\, write the vector \,\textbf{v}\, in terms of \,\textbf{i}\, and \,\textbf{j}.\,

Show Solution

\,7\textbf{i}-3\textbf{j}\,

For the following exercises, use the vectors u = i + 5j, v = −2i− 3j,  and         w = 4i − j.

Find u + (v − w)

Find 4v + 2u

Show Solution

\,-6\textbf{i}-2\textbf{j}\,

For the following exercises, use the given vectors to compute u + v, u − v, and 2u − 3v.

\textbf{u}=\big\langle 2,-3\big\rangle,\textbf{v}=\big\langle 1,5\big\rangle

\textbf{u}=\big\langle -3,4\big\rangle,\textbf{v}=\big\langle -2,1\big\rangle

Show Solution

\textbf{u}+\textbf{v}=\big\langle -5,5\big\rangle,\textbf{u}-\textbf{v}=\big\langle -1,3\big\rangle,2\textbf{u}-3\textbf{v}=\big\langle 0,5\big\rangle

Let v = −4i + 3j. Find a vector that is half the length and points in the same direction as \,\textbf{v}.

Let v = 5i + 2j. Find a vector that is twice the length and points in the opposite direction as \,\textbf{v}.

Show Solution

-10 \textbf{i}-4 \textbf{j}

For the following exercises, find a unit vector in the same direction as the given vector.

a = 3i + 4j

b = −2i + 5j

Show Solution

-\frac{2\sqrt{29}}{29}\textbf{i}+\frac{5\sqrt{29}}{29}\textbf{j}

c = 10i – j

d=-\frac{1}{3}i + \frac{5}{2}j

Show Solution

-\frac{2\sqrt{229}}{229}\textbf{i}+\frac{15\sqrt{229}}{229}\textbf{j}

u = 100i + 200j

u = −14i + 2j

Show Solution

-\frac{7\sqrt{2}}{10}\textbf{i}+\frac{\sqrt{2}}{10}\textbf{j}

For the following exercises, find the magnitude and direction of the vector,\,0\le \theta <2\pi .

\big\langle 0,4\big\rangle

\big\langle 6,5\big\rangle

Show Solution

|v|=7.810,\theta =39.806^{\circ }

\big\langle 2,-5\big\rangle

\big\langle -4,-6\big\rangle

Show Solution

|v|=7.211,\theta =236.310^{\circ }

Given u = 3i − 4j and v = −2i + 3j, calculate \,\textbf{u}\cdot \textbf{v}.

Given u = −i − j and v = i + 5j, calculate \,\textbf{u}\cdot \textbf{v}.

Show Solution

-6

Given \,\textbf{u}=\big\langle -2,4\big\rangle\, and\,\textbf{v}=\big\langle -3,1\big\rangle,\, calculate \,\textbf{u}\cdot \textbf{v}.

Given u =\big\langle -1,6\big\rangle and v=\big\langle 6,-1\big\rangle, calculate \,\textbf{u}\cdot \textbf{v}.

Show Solution

-12

Graphical

For the following exercises, given \,\textbf{v},\, draw \textbf{v}, 3v and \,\frac{1}{2}\,\textbf{v}.

〈2,-1〉

〈-1,4〉
Show Solution

〈-3,-2〉

For the following exercises, use the vectors shown to sketch u + v, u − v, and 2u.

Plot of vectors u and v extending from the same origin point. In terms of that point, u goes to (1,1) and v goes to (-1,2).
Show Solution

Plot of u+v, u-v, and 2u based on the above vectors. In relation to the same origin point, u+v goes to (0,3), u-v goes to (2,-1), and 2u goes to (2,2).

Plot of vectors u and v extending from the same origin point. In terms of that point, u goes to (1,2) and v goes to (1,-1).
Plot of vectors u and v located head to tail. Take u's start point as the origin. In terms of that, u goes from the origin to (3,-2), and v goes from (3,-2) to (2,-3)
Show Solution

Plot of vectors u+v, u-v, and 2u based on the above vectors.Given that u's start point was the origin, u+v starts at the origin and goes to (2,-3); u-v starts at the origin and goes to (4,-1); 2u goes from the origin to (6,-4).

For the following exercises, use the vectors shown to sketch 2u + v.

Plot of the vectors u and v extending from the same point. Taking that base point as the origin, u goes from the origin to (3,1) and v goes from the origin to (2,-2).
Plot of the vectors u and v extending from the same point. Taking that base point as the origin, u goes from the origin to (1,-2) and v goes from the origin to (-3,-2).
Show Solution

Plot of a single vector. Taking the start point of the vector as (0,0) from the above set up, the vector goes from the origin to (-1,-6).

For the following exercises, use the vectors shown to sketch u − 3v.

Plot of the vectors u and v extending from the same point. Taking that base point as the origin, u goes from the origin to (-4,0) and v goes from the origin to (1,-1).
Plot of the vectors u and v extending from the same point. Taking that base point as the origin, u goes from the origin to (1,2) and v goes from the origin to (-2,1).
Show Solution

Vector extending from the origin to (7,5), taking the base as the origin.

For the following exercises, write the vector shown in component form.

Vector going from the origin to (-4,2).
Insert figure(table) alt text: Vector going from the origin to (4,1).
Show Solution

〈4,1〉

Given initial point \,{P}_{1}=\left(2,1\right)\, and terminal point \,{P}_{2}=\left(-1,2\right),\, write the vector \,\textbf{v}\, in terms of \,i\, and \,j,\, then draw the vector on the graph.

Given initial point \,{P}_{1}=\left(4,-1\right)\, and terminal point \,{P}_{2}=\left(-3,2\right),\, write the vector \,\textbf{v}\, in terms of \,\textbf{i}\, and \,\textbf{j}.\, Draw the points and the vector on the graph.

Show Solution

v=-7\textbf{i}+3\textbf{j}

Vector going from (4,-1) to (-3,2).

Given initial point \,{P}_{1}=\left(3,3\right)\, and terminal point \,{P}_{2}=\left(-3,3\right),\, write the vector \,\textbf{v}\, in terms of \,\textbf{i}\, and \,\textbf{j}.\, Draw the points and the vector on the graph.

Extensions

For the following exercises, use the given magnitude and direction in standard position, write the vector in component form.

|v|=6,\theta =45^{\circ }

Show Solution

3\sqrt{2}i+3\sqrt{2}j

|v|=8,\theta =220^{\circ }

|v|=2,\theta =300^{\circ }

Show Solution

i-\sqrt{3}j

|v|=5,\theta =135^{\circ }

A 60-pound box is resting on a ramp that is inclined 12°. Rounding to the nearest tenth,

  1. Find the magnitude of the normal (perpendicular) component of the force.
  2. Find the magnitude of the component of the force that is parallel to the ramp.
Show Solution

a. 58.7; b. 12.5

A 25-pound box is resting on a ramp that is inclined 8°. Rounding to the nearest tenth,

  1. Find the magnitude of the normal (perpendicular) component of the force.
  2. Find the magnitude of the component of the force that is parallel to the ramp.

Find the magnitude of the horizontal and vertical components of a vector with magnitude 8 pounds pointed in a direction of 27° above the horizontal. Round to the nearest hundredth.

Show Solution

x=7.13\, pounds, \,y=3.63\, pounds

Find the magnitude of the horizontal and vertical components of the vector with magnitude 4 pounds pointed in a direction of 127° above the horizontal. Round to the nearest hundredth.

Find the magnitude of the horizontal and vertical components of a vector with magnitude 5 pounds pointed in a direction of 55° above the horizontal. Round to the nearest hundredth.

Show Solution

x=2.87\, pounds, \,y=4.10\,pounds

Find the magnitude of the horizontal and vertical components of the vector with magnitude 1 pound pointed in a direction of 8° above the horizontal. Round to the nearest hundredth.

Real-World Applications

A woman leaves home and walks 3 miles west, then 2 miles southwest. How far from home is she, and in what direction must she walk to head directly home?

Show Solution

4.635 miles, 17.764° N of E

A boat leaves the marina and sails 6 miles north, then 2 miles northeast. How far from the marina is the boat, and in what direction must it sail to head directly back to the marina?

A man starts walking from home and walks 4 miles east, 2 miles southeast, 5 miles south, 4 miles southwest, and 2 miles east. How far has he walked? If he walked straight home, how far would he have to walk?

Show Solution

17 miles. 10.318 miles

A woman starts walking from home and walks 4 miles east, 7 miles southeast, 6 miles south, 5 miles southwest, and 3 miles east. How far has she walked? If she walked straight home, how far would she have to walk?

A man starts walking from home and walks 3 miles at 20° north of west, then 5 miles at 10° west of south, then 4 miles at 15° north of east. If he walked straight home, how far would he have to the walk, and in what direction?

Show Solution

Distance: 2.868. Direction: 86.474° North of West, or 3.526° West of North

A woman starts walking from home and walks 6 miles at 40° north of east, then 2 miles at 15° east of south, then 5 miles at 30° south of west. If she walked straight home, how far would she have to walk, and in what direction?

An airplane is heading north at an airspeed of 600 km/hr, but there is a wind blowing from the southwest at 80 km/hr. How many degrees off course will the plane end up flying, and what is the plane’s speed relative to the ground?

Show Solution

4.924°. 659 km/hr

An airplane is heading north at an airspeed of 500 km/hr, but there is a wind blowing from the northwest at 50 km/hr. How many degrees off course will the plane end up flying, and what is the plane’s speed relative to the ground?

An airplane needs to head due north, but there is a wind blowing from the southwest at 60 km/hr. The plane flies with an airspeed of 550 km/hr. To end up flying due north, how many degrees west of north will the pilot need to fly the plane?

Show Solution

4.424°

An airplane needs to head due north, but there is a wind blowing from the northwest at 80 km/hr. The plane flies with an airspeed of 500 km/hr. To end up flying due north, how many degrees west of north will the pilot need to fly the plane?

As part of a video game, the point\,\left(5,7\right)\,is rotated counterclockwise about the origin through an angle of 35°. Find the new coordinates of this point.

Show Solution

\left(0.081,8.602\right)

As part of a video game, the point\,\left(7,3\right)\,is rotated counterclockwise about the origin through an angle of 40°. Find the new coordinates of this point.

Two children are throwing a ball back and forth straight across the back seat of a car. The ball is being thrown 10 mph relative to the car, and the car is traveling 25 mph down the road. If one child doesn’t catch the ball, and it flies out the window, in what direction does the ball fly (ignoring wind resistance)?

Show Solution

21.801°, relative to the car’s forward direction

Two children are throwing a ball back and forth straight across the back seat of a car. The ball is being thrown 8 mph relative to the car, and the car is traveling 45 mph down the road. If one child doesn’t catch the ball, and it flies out the window, in what direction does the ball fly (ignoring wind resistance)?

A 50-pound object rests on a ramp that is inclined 19°. Find the magnitude of the components of the force parallel to and perpendicular to (normal) the ramp to the nearest tenth of a pound.

Show Solution

parallel: 16.28, perpendicular: 47.28 pounds

Suppose a body has a force of 10 pounds acting on it to the right, 25 pounds acting on it upward, and 5 pounds acting on it 45° from the horizontal. What single force is the resultant force acting on the body?

Suppose a body has a force of 10 pounds acting on it to the right, 25 pounds acting on it ─135° from the horizontal, and 5 pounds acting on it directed 150° from the horizontal. What single force is the resultant force acting on the body?

Show Solution

19.35 pounds, 231.54° from the horizontal

The condition of equilibrium is when the sum of the forces acting on a body is the zero vector. Suppose a body has a force of 2 pounds acting on it to the right, 5 pounds acting on it upward, and 3 pounds acting on it 45° from the horizontal. What single force is needed to produce a state of equilibrium on the body?

Suppose a body has a force of 3 pounds acting on it to the left, 4 pounds acting on it upward, and 2 pounds acting on it 30° from the horizontal. What single force is needed to produce a state of equilibrium on the body? Draw the vector.

Show Solution

5.1583 pounds, 75.8° from the horizontal

Chapter Review Exercises

Non-right Triangles: Law of Sines

For the following exercises, assume \,\alpha \, is opposite side\,a,\beta \, is opposite side \,b,\, and \,\gamma \, is opposite side \,c.\, Solve each triangle, if possible. Round each answer to the nearest tenth.

\beta =50^{\circ },a=105,b=45

Show Solution

Not possible

\alpha =43.1^{\circ },a=184.2,b=242.8

Solve the triangle.

Triangle with standard labels. Angle A is 36 degrees with opposite side a unknown. Angle B is 24 degrees with opposite side b = 16. Angle C and side c are unknown.

Show Solution

C=120^{\circ },a=23.1,c=34.1

Find the area of the triangle.

A triangle. One angle is 75 degrees with opposite side unknown. The adjacent sides to the 75 degree angle are 8 and 11.

A pilot is flying over a straight highway. He determines the angles of depression to two mileposts, 2.1 km apart, to be 25° and 49°, as shown in (Figure 20). Find the distance of the plane from point \,A\, and the elevation of the plane.

Diagram of a plane flying over a highway. It is to the left and above points A and B on the ground in that order. There is a horizontal line going through the plan parallel to the ground. The angle formed by the horizontal line, the plane, and the line from the plane to point B is 25 degrees. The angle formed by the horizontal line, the plane, and point A is 49 degrees.
Figure 20.
Show Solution

distance of the plane from point \,A:\, 2.2 km, elevation of the plane: 1.6 km

Non-right Triangles: Law of Cosines

Solve the triangle, rounding to the nearest tenth, assuming \,\alpha \, is opposite side \,a,\beta \, is opposite side \,b,\, and \,\gamma \, is opposite side c:\,a=4, b=6,c=8.

Solve the triangle in (Figure 21), rounding to the nearest tenth.

A standardly labeled triangle. Angle A is 54 degrees with opposite side a unknown. Angle B is unknown with opposite side b=15. Angle C is unknown with opposite side C=13.
Figure 21.
Show Solution

B=71.0^{\circ },C=55.0^{\circ },a=12.8

Find the area of a triangle with sides of length 8.3, 6.6, and 9.1.

To find the distance between two cities, a satellite calculates the distances and angle shown in (Figure 22) (not to scale). Find the distance between the cities. Round answers to the nearest tenth.

Diagram of a satellite above and to the right of two cities. The distance from the satellite to the closer city is 210 km. The distance from the satellite to the further city is 250 km. The angle formed by the closer city, the satellite, and the other city is 1.8 degrees.
Figure 22.
Show Solution

40.6 km

Polar Coordinates

Plot the point with polar coordinates \,\left(3,\frac{\pi }{6}\right).

Plot the point with polar coordinates \,\left(5,-\frac{2\pi }{3}\right)

Show Solution

Polar coordinate grid with a point plotted on the fifth concentric circle 2/3 the way between pi and 3pi/2 (closer to 3pi/2).

Convert \,\left(6,-\frac{3\pi }{4}\right)\, to rectangular coordinates.

Convert \,\left(-2,\frac{3\pi }{2}\right)\, to rectangular coordinates.

Show Solution

\,\left(0,2\right)\,

Convert \left(7,-2\right) to polar coordinates.

Convert \left(-9,-4\right)
to polar coordinates.

Show Solution

\left(9.8489,203.96°\right)

For the following exercises, convert the given Cartesian equation to a polar equation.

x=-2

{x}^{2}+{y}^{2}=64

Show Solution

r=8

{x}^{2}+{y}^{2}=-2y

For the following exercises, convert the given polar equation to a Cartesian equation.

r=7\text{cos}\,\theta

Show Solution

{x}^{2}+{y}^{2}=7x

r=\frac{-2}{4\mathrm{cos}\,\theta +\mathrm{sin}\,\theta }

For the following exercises, convert to rectangular form and graph.

\theta =\frac{3\pi }{4}

Show Solution

y=-x

Plot of the function y=-x in rectangular coordinates.

r=5\mathrm{sec}\,\theta

Polar Coordinates: Graphs

For the following exercises, test each equation for symmetry.

r=4+4\mathrm{sin}\,\theta
Show Solution

symmetric with respect to the line\theta =\frac{\pi }{2}

r=7

Sketch a graph of the polar equation\,r=1-5\mathrm{sin}\,\theta .\,Label the axis intercepts.

Show Solution

Graph of the given polar equation - an inner loop limaçon.

Sketch a graph of the polar equation\,r=5\mathrm{sin}\left(7\theta \right).

Sketch a graph of the polar equation\,r=3-3\mathrm{cos}\,\theta

Show Solution

Graph of the given polar equation - a cardioid.

Polar Form of Complex Numbers

For the following exercises, find the absolute value of each complex number.

-2+6i

4-\text{​}3i

Show Solution

5

Write the complex number in polar form.

5+9i

\frac{1}{2}-\frac{\sqrt{3}}{2}\text{​}i

Show Solution

\mathrm{cis}\left(-\frac{\pi }{3}\right)

For the following exercises, convert the complex number from polar to rectangular form.

z=5\mathrm{cis}\left(\frac{5\pi }{6}\right)

z=3\mathrm{cis}\left(40°\right)

Show Solution

2.3+1.9i

For the following exercises, find the product\,{z}_{1}{z}_{2}\,in polar form.

{z}_{1}=2\mathrm{cis}\left(89°\right)

{z}_{2}=5\mathrm{cis}\left(23°\right)

{z}_{1}=10\mathrm{cis}\left(\frac{\pi }{6}\right)

{z}_{2}=6\mathrm{cis}\left(\frac{\pi }{3}\right)

Show Solution

60\mathrm{cis}\left(\frac{\pi }{2}\right)

For the following exercises, find the quotient\,\frac{{z}_{1}}{{z}_{2}}\,in polar form.

{z}_{1}=12\mathrm{cis}\left(55°\right)

{z}_{2}=3\mathrm{cis}\left(18°\right)

{z}_{1}=27\mathrm{cis}\left(\frac{5\pi }{3}\right)

{z}_{2}=9\mathrm{cis}\left(\frac{\pi }{3}\right)

Show Solution

3\mathrm{cis}\left(\frac{4\pi }{3}\right)

For the following exercises, find the powers of each complex number in polar form.

Find\,{z}^{4}\,when\,z=2\mathrm{cis}\left(70°\right)

Find\,{z}^{2}\,when\,z=5\mathrm{cis}\left(\frac{3\pi }{4}\right)

Show Solution

25\mathrm{cis}\left(\frac{3\pi }{2}\right)

For the following exercises, evaluate each root.

Evaluate the cube root of\,z\,when\,z=64\mathrm{cis}\left(210°\right).

Evaluate the square root of\,z\,when\,z=25\mathrm{cis}\left(\frac{3\pi }{2}\right).

Show Solution

5\mathrm{cis}\left(\frac{3\pi }{4}\right),5\mathrm{cis}\left(\frac{7\pi }{4}\right)

For the following exercises, plot the complex number in the complex plane.

6-2i

-1+3i

Show Solution

Plot of -1 + 3i in the complex plane (-1 along the real axis, 3 along the imaginary).

Parametric Equations

For the following exercises, eliminate the parameter\,t\,to rewrite the parametric equation as a Cartesian equation.

\{\begin{array}{l}x(t)=3t-1\hfill \\ y(t)=\sqrt{t}\hfill \end{array}

\{\begin{array}{l}x(t)=-\mathrm{cos}\,t\hfill \\ y(t)=2{\mathrm{sin}}^{2}t \hfill \end{array}

Show Solution

{x}^{2}+\frac{1}{2}y=1

Parameterize (write a parametric equation for) each Cartesian equation by using\,x\left(t\right)=a\mathrm{cos}\,t\,and\,y\left(t\right)=b\mathrm{sin}\,t\,for\,\frac{{x}^{2}}{25}+\frac{{y}^{2}}{16}=1.

Parameterize the line from\,\left(-2,3\right)\,to\,\left(4,7\right)\,so that the line is at\,\left(-2,3\right)\,at\,t=0\,and\,\left(4,7\right)\,at\,t=1.

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\{\begin{array}{l}x(t)=-2+6t\hfill \\ y(t)=3+4t\hfill \end{array}

Parametric Equations: Graphs

For the following exercises, make a table of values for each set of parametric equations, graph the equations, and include an orientation; then write the Cartesian equation.

\{\begin{array}{l}x(t)=3{t}^{2}\hfill \\ y(t)=2t-1\hfill \end{array}

\{\begin{array}{l}x(t)={e}^{t}\hfill \\ y(t)=-2{e}^{5\,t}\hfill \end{array}

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y=-2{x}^{5}

Plot of the given parametric equations.

\{\begin{array}{l}x(t)=3\mathrm{cos}\,t\hfill \\ y(t)=2\mathrm{sin}\,t\hfill \end{array}

A ball is launched with an initial velocity of 80 feet per second at an angle of 40° to the horizontal. The ball is released at a height of 4 feet above the ground.

  1. Find the parametric equations to model the path of the ball.
  2. Where is the ball after 3 seconds?
  3. How long is the ball in the air?
Show Solution
  1. \{\begin{array}{l}x(t)=(80\mathrm{cos}(40°))t\\ y(t)=-16{t}^{2}+(80\mathrm{sin}(40°))t+4\end{array}
  2. The ball is 14 feet high and 184 feet from where it was launched.
  3. 3.3 seconds

Vectors

For the following exercises, determine whether the two vectors,\,u\,and\,v,\,are equal, where\,u\,has an initial point\,{P}_{1}\,and a terminal point\,{P}_{2},\,and\,v\,has an initial point\,{P}_{3}\,and a terminal point\,{P}_{4}.

{P}_{1}=\left(-1,4\right),{P}_{2}=\left(3,1\right),{P}_{3}=\left(5,5\right)and\,{P}_{4}=\left(9,2\right)

{P}_{1}=\left(6,11\right),{P}_{2}=\left(-2,8\right),{P}_{3}=\left(0,-1\right)\,and\,{P}_{4}=\left(-8,2\right)

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not equal

For the following exercises, use the vectors\,u=2i-j\text{,}v=4i-3j\text{,}\,and\,w=-2i+5j\,to evaluate the expression.

u − v

2v − u + w

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4i

For the following exercises, find a unit vector in the same direction as the given vector.

a = 8i − 6j

b = −3i − j

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-\frac{3\sqrt{10}}{10}i-\frac{\sqrt{10}}{10}j

For the following exercises, find the magnitude and direction of the vector.

〈6,-2〉

〈-3,-3〉

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Magnitude:\,3\sqrt{2},\,Direction:\text{225°}

For the following exercises, calculate\,u\cdot v\text{.}

u = −2i + j and v = 3i + 7j

u = i + 4j and v = 4i + 3j

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\text{16}

Given v=〈-3,4〉draw v, 2v, and \,\frac{1}{2}v.

Given the vectors shown in (Figure), sketch u + v, u − v and 3v.

Diagram of vectors v, 2v, and 1/2 v. The 2v vector is in the same direction as v but has twice the magnitude. The 1/2 v vector is in the same direction as v but has half the magnitude.
Figure 23.
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Diagram of vectors u and v. Taking u's starting point as the origin, u goes from the origin to (4,1), and v goes from (4,1) to (6,0).

Given initial point\,{P}_{1}=\left(3,2\right)\,and terminal point\,{P}_{2}=\left(-5,-1\right),\,write the vector\,v\,in terms of\,\,i\,\,and\,j.\,Draw the points and the vector on the graph.

Practice Test

Assume\,\alpha \,is opposite side\,a,\beta \,is opposite side\,b,\,and\,\gamma \,is opposite side\,c.\,Solve the triangle, if possible, and round each answer to the nearest tenth, given\,\beta =68°,b=21,c=16.

\alpha =67.1°,\gamma =44.9°,a=20.9

Find the area of the triangle in (Figure). Round each answer to the nearest tenth.

A triangle. One angle is 60 degrees with opposite side 6.25. The other two sides are 5 and 7.
Figure 24.

A pilot flies in a straight path for 2 hours. He then makes a course correction, heading 15° to the right of his original course, and flies 1 hour in the new direction. If he maintains a constant speed of 575 miles per hour, how far is he from his starting position?

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\text{1712 miles}

Convert\,\left(2,2\right)\,
to polar coordinates, and then plot the point.

Convert\,\left(2,\frac{\pi }{3}\right)\,to rectangular coordinates.

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\left(1,\sqrt{3}\right)

Convert the polar equation to a Cartesian equation:\,{x}^{2}+{y}^{2}=5\mathrm{y.}

Convert to rectangular form and graph:r=-3\mathrm{csc}\,\theta .

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y=-3

Plot of the given equation in rectangular form - line y=-3.

Test the equation for symmetry:\,r=-4\mathrm{sin}\left(2\theta \right).

Graph\,r=3+3\mathrm{cos}\,\theta .

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Graph of the given equations - a cardioid.

Graph\,r=3-5\text{sin}\,\theta .

Find the absolute value of the complex number
5-9i.

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\sqrt{106}

Write the complex number in polar form:\,4+i\text{.}

Convert the complex number from polar to rectangular form:\,z=5\text{cis}\left(\frac{2\pi }{3}\right).

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\frac{-5}{2}+i\frac{5\sqrt{3}}{2}

Given\,{z}_{1}=8\mathrm{cis}\left(36°\right)\,and\,{z}_{2}=2\mathrm{cis}\left(15°\right),evaluate each expression.

{z}_{1}{z}_{2}

\frac{{z}_{1}}{{z}_{2}}

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4\mathrm{cis}\left(21°\right)

{\left({z}_{2}\right)}^{3}

\sqrt{{z}_{1}}

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2\sqrt{2}\mathrm{cis}\left(18°\right),2\sqrt{2}\mathrm{cis}\left(198°\right)

Plot the complex number\,-5-i\,in the complex plane.

Eliminate the parameter\,t\,to rewrite the following parametric equations as a Cartesian equation: \,\{\begin{array}{l}x(t)=t+1\hfill \\ y(t)=2{t}^{2}\hfill \end{array}.

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y=2{\left(x-1\right)}^{2}

Parameterize (write a parametric equation for) the following Cartesian equation by using\,x\left(t\right)=a\mathrm{cos}\,t\,and\,y\left(t\right)=b\mathrm{sin}\,t:[latex]\frac{{x}^{2}}{36}+\frac{{y}^{2}}{100}=1.[/latex]

Graph the set of parametric equations and find the Cartesian equation:\,\{\begin{array}{l}x(t)=-2\mathrm{sin}\,t\hfill \\ y(t)=5\mathrm{cos}\,t\hfill \end{array}.

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Graph of the given equations - a vertical ellipse.

A ball is launched with an initial velocity of 95 feet per second at an angle of 52° to the horizontal. The ball is released at a height of 3.5 feet above the ground.

  1. Find the parametric equations to model the path of the ball.
  2. Where is the ball after 2 seconds?
  3. How long is the ball in the air?

For the following exercises, use the vectors u = i − 3j and v = 2i + 3j.

Find 2u − 3v.

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−4i − 15j

Calculate\,u\cdot v.

Find a unit vector in the same direction as\,v.

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\frac{2\sqrt{13}}{13}i+\frac{3\sqrt{13}}{13}j

Given vector\,v\,has an initial point\,{P}_{1}=\left(2,2\right)\,and terminal point\,{P}_{2}=\left(-1,0\right),\,write the vector\,v\,in terms of\,i\,and\,j.\,On the graph, draw\,v,\,and\,-v.\,

Glossary

dot product
given two vectors, the sum of the product of the horizontal components and the product of the vertical components
initial point
the origin of a vector
magnitude
the length of a vector; may represent a quantity such as speed, and is calculated using the Pythagorean Theorem
resultant
a vector that results from addition or subtraction of two vectors, or from scalar multiplication
scalar
a quantity associated with magnitude but not direction; a constant
scalar multiplication
the product of a constant and each component of a vector
standard position
the placement of a vector with the initial point at\,\left(0,0\right)\,and the terminal point\,\left(a,b\right),\,represented by the change in the x-coordinates and the change in the y-coordinates of the original vector
terminal point
the end point of a vector, usually represented by an arrow indicating its direction
unit vector
a vector that begins at the origin and has magnitude of 1; the horizontal unit vector runs along the x-axis and is defined as\,{v}_{1}=〈1,0〉\,the vertical unit vector runs along the y-axis and is defined as\,{v}_{2}=〈0,1〉.
vector
a quantity associated with both magnitude and direction, represented as a directed line segment with a starting point (initial point) and an end point (terminal point)
vector addition
the sum of two vectors, found by adding corresponding components

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